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Math Help - Can anyone help me with this Trig Identity double angle question?

  1. #1
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    Can anyone help me with this Trig Identity double angle question?

    Hi guys I am having real difficulty cracking this tri identity.

    square root 2 sin (theta- pi/4) = sin theta - cos theta

    I have attempted it ever so slightly but am struggling at transposing the LHS to match the RHS

    pi/4 I figured was square root 2/2 as a trigonometric value, I thought that this would have some influenece in cancelling out the initial square root of 2.

    Another thought was to calculate the pi/4 = 0.785 and also square root of 2 sin = 1.414.

    Again I am not entirely sure which path or direction to take with this one, any help would be massively appreciated.

    Thanks in advance
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  2. #2
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    \sin\left(\alpha-\beta\right) = \sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}
    and \sin\frac{\pi}{4} = \cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}}. That should do it.
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  3. #3
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    Thanks Coffee machine,

    Very kind, please can you explain how you transposed to get to the result that you did, I was confused about how I can get rid of the initial square root of 2 (space) sin?,...... crikey I have been banging my head on this one for over a week.

    Much appreciated
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  4. #4
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    My suggestion was to use the addition identity:

    \sin\left(\alpha-\beta\right) = \sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}.

    You have:

    \sin\left(\theta-\frac{\pi}{4}\right) = \sin{\theta}\cos{\frac{\pi}{4}}-\cos{\theta}\sin{\frac{\pi}{4}}.

    The value of \cos\frac{\pi}{4} is \frac{1}{\sqrt{2}} and that of \sin\frac{\pi}{4} is \frac{1}{\sqrt{2}}, so we have:

    \sin\left(\theta-\frac{\pi}{4}\right) = \sin{\theta}\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\sin{\theta} = \frac{1}{\sqrt{2}}\left(\sin{\theta}-\cos{\theta}\right).

    Thus \sqrt{2}\sin\left(\theta-\frac{\pi}{4}\right) =  \frac{\sqrt{2}}{\sqrt{2}}\left(\sin{\theta}-\cos{\theta}\right) = \sin{\theta}-\cos{\theta}.
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  5. #5
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    Very kind, thanks ever so much, I can see how this progresses now.
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